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MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: Xinyu Jing on October 07, 2019, 09:26:17 AM

Title: QUIZ2 TUT 0204
Post by: Xinyu Jing on October 07, 2019, 09:26:17 AM
Question: (𝑥+2)𝑠𝑖𝑛(𝑦)+𝑥𝑐𝑜𝑠(𝑦)𝑦′=0   𝑢=𝑥𝑒𝑥
Solution:
𝑀=(𝑥+2)𝑠𝑖𝑛(𝑦)    𝑁=𝑥𝑐𝑜𝑠(𝑦)
𝑀𝑦=(𝑥+2)𝑐𝑜𝑠(𝑦)     𝑁𝑥=𝑐𝑜𝑠(𝑦)
therefore 𝑀𝑦≠𝑁𝑥 , therefore the equation is not exact.
multiplies the given integrating factor
(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)𝑦′=0
𝑀=(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)  𝑁=𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)
𝑀𝑦=(𝑥+2)𝑥𝑒𝑥𝑐𝑜𝑠(𝑦)  𝑁𝑥=𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦) + 2𝑥𝑒𝑥𝑐𝑜𝑠(𝑦)
then 𝑀𝑦=𝑁𝑥,  therefore it becomes exact.
𝜙(𝑥,𝑦) s.t 𝜙𝑥=𝑀   𝜙𝑦=𝑁
𝜙=∫𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)𝑑𝑦
𝜙=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ(𝑥)
𝜙𝑥=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦) + 2𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ′(𝑥)=𝑀=(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+0
ℎ′(𝑥)=0
ℎ(𝑥)=𝑐
𝜙(𝑥,𝑦)=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ(𝑥)=𝑐